@NorbertSchuch I asked about it because I was looking for a lower bound to the decay of the l.h.s. (the Hilbert-Schmidt inner product of $\rho(0)$ with $\rho(t)$) via the the r.h.s.. This because the exponent in the r.h.s. $\langle \mathcal{L}\rangle_{\rho}$ might vanish in the thermodynamic limit for some special states $\rho$ that are in some way ''continuously connected'' to a (pure) stationary state in the large-size limit, thus freezing the l.h.s. to 1.

Thank you for the very clean answer! It would be nice to know if it holds on all states but the stationary ones, if you have any idea about that I can open a new question with this twist to the problem. Have a nice day!

Thanks for the link! The problem here is that in the case of a Lindbladian superoperator, the expectation value is not real. So does Jensen's inequality hold with the absolute value on both sides for operators with complex expectation values?

I edited the notation above. I am asking whether the above inequality holds for an arbitrary density matrix $\rho$ over which the average is taken. As for the inequality sign, I guess that the problem is that on both sides one has complex numbers; then I would put the absolute value on both sides.

I would define $\langle A\rangle_{\rho}:= Tr[\rho A(\rho)]=(\rho, A(\rho))$, where the scalar product (•,•) is defined as $(A,B)=Tr[A^{+} B]$.